Lightweight Electric/Hybrid Vehicle Design

Design for optimum body-structural and running-gear performance efficiency 213
known as tension coefficients is applied. This is based on the fact that proportionality exists between
both resolved components of length and of force. A tension coefficient
S/L = F x /X = F y /Y = F z /Z
for member force S, in a length L, has projecting force vectors and length components, F x , F y , F z
and X,Y,Z on perpendicular coordinate axes.
The view at (d) shows one bay of a space truss, that might be an extension used to support an
engine/gearbox unit behind a monocoque bodyshell structure. For analysis, the frame is assumed
to have rigid plates at b 1,2,3,4 and c 1,2,3,4 which offer no force reaction perpendicular to their own
planes (zero axial warping constraint). When considering the torsional load case, for the bay, b 1 c 1
and b 4 c 4 member forces are zero by resolving at b 1 and c 4 , observing zero axial constraint of the
truss. Remaining members (the envelope) have equal force components F 2 at the bulkheads and
tension coefficients for all of them are F z /Z. This common value can be determined by projecting
envelope member forces onto the bulkhead planes, as shown. By taking moments about any axis
O, for any one member (c 3 b 4 , say, with projected length d), contribution to torque reaction is = rtd
(half the area of the triangle formed by joining O to c 3 and b 4 ). Thus total torque is 2tA, by summation.
8.4.1 STIFF-JOINTED FRAMES
Where structural members are curved or cranked over wheel arches or drive shaft tunnels the unit
load method of analysis applied to stiff-jointed frames is particularly useful, Fig. 8.8. It is best
understood by considering a small elastic element in a curved beam (a) which is otherwise assumed
to be rigid. An imaginary load w, at P, is then considered to cause vertical displacement Δ; the
beam deflects under the influence of the bending moment and it can be shown that Δ = ∫ Mm/
EI . ds where m is bending moment due to imaginary unit load at P and M due to the real external
load system. An example might be a car floor transverse crossbearer having a ‘tunnel’ portion for
exhaust-system and prop-shaft clearance, (b). To find the downward deflection at B, the imaginary
unit load is applied at that point and it develops reactions of two-thirds at A and one-third at D.
The bending moments for the real and imaginary loads are shown at (c) which also illustrates the
subsequent calculation to obtain the deflection due to the driver and seat represented by distributed
load w over length l. Another example is the part of the frame at (d) subject to twisting and
bending, deflection due to twist = ∫ Tt dx/GJ. This deflection would then be added to that due to
bending, obtained from the formula given in the earlier section. Usually the deflection due to axial
straining of the elements is so small that it can be discounted.
A good example of a frame subject to twisting and bending is the portal shape frame at (e)
having loading and support, in the horizontal plane, indicated. If it is required to find the vertical
deflection at D for the load P applied at B, equating vertical forces and moments is first carried out
to find the reactions at the supports. Next step is to break the frame up into its elements and
determine the bending moments in each – ensuring compatibility of these, and that associated
loads are ‘transferred’ across the artificially broken joints. Integrations have to be carried out for
the three deflection modes as follows: ∫ Sb . ds/AE + ∫ Mm . ds/EI + ∫ Ff . ds/AG to obtain the
combined deflection. Both vertical and horizontal components can be obtained by applying vertical
and horizontal imaginary unit loads, in turn, at the point where the deflection is to be determined.
For the example shown in the figure, elastic and shear moduli, E and G, are 200 and 80 GN/m 2
respectively. S is axial load and F the shear load, the latter being negative in compression. The
values of these, with the bending moments, are as shown at (f right). These can readily be substituted
in the expression (f left) to determine deflections as a factor of P.